Integrals

1. Integrals 5

2. The Substitution Rule 5.5

3. The Substitution Rule • Because of the Fundamental Theorem, it’s important to be able to find antiderivatives. • But our antidifferentiation formulas don’t tell us how to evaluate integrals such as • To find this integral we use a new variable; we change from the variable x to a new variable u.

4. The Substitution Rule • Suppose that we let u be the quantity under the root sign in (1): u = 1 + x2. Then the differential of u is du = 2x dx. • Notice that if the dx in the notation for an integral were to be interpreted as a differential, then the differential 2x dx would occur in (1) and so, formally, without justifying our calculation, we could write

5. The Substitution Rule • But now we can check that we have the correct answer by using the Chain Rule to differentiate the final function of Equation 2: • In general, this method works whenever we have an integral that we can write in the form f(g(x))g(x) dx.

6. The Substitution Rule • Observe that if F = f,then • F(g(x))g(x) dx = F(g(x)) + C • because, by the Chain Rule, • [F(g(x))] = F(g(x))g(x) • If we make the “change of variable” or “substitution” u = g(x), • then from Equation 3 we have • F(g(x))g(x) dx = F(g(x)) + C = F(u) + C = F(u) du • or, writing F = f, we get • f(g(x))g(x) dx =f(u) du

7. The Substitution Rule • Thus we have proved the following rule. • Notice that the Substitution Rule for integration was proved using the Chain Rule for differentiation. • Notice also that if u = g(x), then du = g(x) dx, so a way to remember the Substitution Rule is to think of dx and du in (4) as differentials.

8. The Substitution Rule • Thus the Substitution Rule says: It is permissible to operate with dx and du after integral signs as if they were differentials.

9. Example 1 – Using the Substitution Rule • Find x3 cos(x4 + 2) dx. • Solution: • We make the substitution u = x4 + 2 because its differential is du = 4x3 dx, which, apart from the constant factor 4, occurs in the integral. • Thus, using x3 dx =du and the Substitution Rule, we have • x3 cos(x4 + 2) dx = cos udu • =  cos u du

10. Example 1 – Solution cont’d • = sin u + C • = sin(x4 + 2)+ C • Notice that at the final stage we had to return to the original variable x.

11. Definite Integrals

12. Definite Integrals • When evaluating a definite integral by substitution, two methods are possible. One method is to evaluate the indefinite integral first and then use the Evaluation Theorem. • For example, • Another method, which is usually preferable, is to change the limits of integration when the variable is changed.

13. Definite Integrals

14. Example 6 – Substitution in a Definite Integral • Evaluate . • Solution: • Let u = 2x + 1. Then du = 2 dx, so dx = du. • To find the new limits of integration we note that • when x = 0, u = 2(0) + 1 = 1 • and • when x = 4, u = 2(4) + 1 = 9 • Therefore

15. Example 6 – Solution cont’d • Observe that when using (5) we do not return to the variable x after integrating. We simply evaluate the expression in u between the appropriate values of u.

16. Symmetry

17. Symmetry • The following theorem uses the Substitution Rule for • Definite Integrals (5) to simplify the calculation of integrals of • functions that possess symmetry properties.

18. Symmetry • Theorem 6 is illustrated by Figure 4. • For the case where f is positive and even, part (a) says that the area under y = f(x) from –a to a is twice the area from 0 to a because of symmetry. Figure 4

19. Symmetry • Recall that an integral can be expressed as the • area above the x-axis and below y = f(x) minus the area • below the axis and above the curve. • Thus part (b) says the integral is 0 because the areas cancel.

20. Example 9 – Integrating an Even Function • Since f(x) = x6 + 1 satisfies f(–x) = f(x), it is even and so

21. Example 10 – Integrating an Odd Function • Since f(x) = (tan x)/(1 + x2 + x4) satisfies f(–x) = –f(x), it is odd and so